Moving Charges And Magnetism PDF

MOVING CHARGES AND MAGNETISM MOVING CHARGES AND MAGNETISM GENERAL KEY CONCEPT 1. Force on a moving charge:– A moving charge is a source of magnetic field. Z...

MOVING CHARGES AND MAGNETISM MOVING CHARGES AND MAGNETISM GENERAL KEY CONCEPT 1. Force on a moving charge:– A moving charge is a source of magnetic field. Z +q v cos B Y  sin v X v   Let a positive charge q is moving in a uniform magnetic field B with velocity v.’ F  q  F  v sin  F  B  F  qBv sin F = kq Bv sin  [k = constant] k = 1 in S.I. system.     F = qBv sin and F  q( v  B )  2. Magnetic field strength ( B ) : In the equation F = qBv sin  , if q = 1, v = 1, sin = 1 i.e.  = 90° then F = B.  Magnetic field strength is defined as the force experienced by a unit charge moving with unit velocity perpendicular to the direction of magnetic field. Special Cases: (1) It  = 0° or 180°, sin = 0  F=0 A charged particle moving parallel to the magnetic field, will not experience any force. (2) If v = 0, F = 0 A charged particle at rest in a magnetic field will not experience any force. (3) If  = 90°, sin = 1 then the force is maximum Fmax. = qvB A charged particle moving perpendicular to magnetic field will experience maximum force. 3. S.I. unit of magnetic field intensity. It is called tesla (T). F B qv sin  If q = 1C, v = 1m/s,  = 90° i.e. sin = 1 and F = 1N Then B = 1T.  MOVING CHARGES AND MAGNETISM The strength of magnetic field at a point is said to be 1T if a charge of 1C while moving at right angle to a magnetic field, with a velocity of 1 m/s experiences a force of 1N at that point. 4. Biot-Savart’s law:– The strength of magnetic field Y or magnetic flux density at a point P (dB) due to current element dl depends on, (i) dB  I dl  r (ii) dB  dl P (iii) dB  sin  X I 1 (iv) dB  , r2 Idl sin  Idl sin  Combining, dB   dB  k [k = Proportionality constant] r2 r2 0 In S.I. units, k  where µ0 is called permeability of free space. 4  0 = 4 × 10–7 TA–1m   0 Idl sin   0 (dl  r )  dB  and dB  I 4 r 2 4 r3    d B is perpendicular to the plane containing d and r and is directed inwards. 5. Applications of Biot-Savart’s law:– (a) Magnetic field (B) at the Centre of a Circular Coil Carrying Current.  nI B 0 2r where n is the number of turns of the coil. I is the current in the coil and r is the radius of the coil. I 2 (b) Magnetic field due to a straight conductor carrying current. a P 1 0 I B (sin 2  sin 1 ) 4a where a is the perpendicular distance of the conductor from the point where field is to the measured. 1 and 2 are the angles made by the two ends of the conductor with the point. (c) For an infinitely long conductor, 1  2   / 2 0 2I  B=. 4 a (d) Magnetic field at a point on the axis of a Circular Coil Carrying Current. when point P lies far away from the centre of the coil. 0 2M B. 3 4 x where M = nIA = magnetic dipole moment of the coil. x is the distance of the point where the field is to the measured, n is the number of turns, I is the current and A is the area of the coil. MOVING CHARGES AND MAGNETISM 6. Ampere’s circuital law:–  The line integral of magnetic field B around any closed path in vacuum is  0   times the total current through the closed path. i.e.  B.d l   0 I 7. Application of Ampere’s circuital law:– (a) Magnetic field due to a current carrying solenoid, B = µ0nI n is the number of turns per unit length of the solenoid. µ0 nI At the edge of a short solenoid, B = 2 (b) Magnetic field due to a toroid or endless solenoid B = µ0nI 8. Motion of a charged particle in uniform electric field:– The path of a charged particle in an electric field is a parabola. 2mv2 Equation of the parabola is x 2  y qE where x is the width of the electric field. y is the displacement of the particle from its straight path. v is the speed of the charged particle. q is the charge of the particle E is the electric field intensity. m is the mass of the particle. 9. Motion of the charged particle in a magnetic field. The path of a charged particle   moving in a uniform magnetic field ( B ) with a velocity v making an angle  with  B is a helix. n v si v  B O cos  The component of velocity v cos  will not provide a force to the charged particle, so under this velocity the particle with move forward with a constant velocity  along the direction of B. The other component v sin  will produce the force F = q Bv sin  , which will supply the necessary centripetal force to the charged particle in moving along a circular path of radius r. m(vsin )2  Centripetal force = = B qv sin  r Bqr  v sin  = m v sin  Bq Angular velocity of rotation = w =  r m  Bq Frequency of rotation =   2 2m 1 2m Time period of revolution = T =  Bq  MOVING CHARGES AND MAGNETISM 10. Cyclotron: It is a device used to accelerate and hence energies the positively charged particle. This is done by placing the particle in an oscillating electric field and a perpendicular magnetic field. The particle moves in a circular path.  Centripetal force = magnetic Lorentz force mv2 mv  = Bqv  = r  radius of the circular path r Bq r m Time to travel a semicircular path =  = constant. v Bq If v0 be the maximum velocity of the particle and r0 be the maximum radius of its path then mv0 2 Bqr0  Bqv0  v0  r0 m 1 1  Bqr0  2 B2 q 2 r0 2 Max. K.E. of the particle = mv 0 2  m   (K.E.)max. = 2 2  m  2m 2 m Time period of the oscillating electric field  T =. Bq Time period is independent of the speed and radius. 1 Bq Cyclotron frequency =   T 2m Bq Cyclotron angular frequency = 0  2  m 11. Force on a current carrying conductor placed in a magnetic field:    F  I   B or F = I  B sin  where I is the current through the conductor B is the magnetic field intensity. l is the length of the conductor.  is the angle between the direction of current and magnetic field. (i) When  = 0° or 180°, sin  = 0  F = 0  When a conductor is placed along the magnetic field, no force will act on the conductor. (ii) When  = 90°, sin  = 1, F is maximum. Fmax = I  B when the conductor is placed perpendicular to the magnetic field, it will experience maximum force. 12. Force between two parallel conductors carrying current:– (a) When the current is in same direction the two conductors will attract each other with a force 0 2I1I2 F. per unit length of the conductor 4 r MOVING CHARGES AND MAGNETISM (b) When the current is in opposite direction the two conductors will repel each other with the same force. (c) S.I. unit of current is 1 ampere. (A). 1A is the current which on flowing through each of the two parallel uniform linear conductor placed in free space at a distance of 1 m from each other produces a force of 2 × 10–7 N/m along their lengths. 13. Torque on a current carrying coil placed in a magnetic field:–      M  B   = MB sin = nIBA sin where M is the magnetic dipole moment of the coil. M = nIA where n is the number of turns of the coil. I is the current through the coil. B is the magnetic field intensity. A is the area of the coil.   is the angle between the magnetic field  B and the perpendicular to the plane of the coil. Special Cases: (i) If the coil is placed parallel to magnetic field  = 0°, cos  = 1 then torque is maximum. max.  nIBA (ii) If the coil is placed perpendicular to magnetic field,  = 90°, cos  = 0   =0 14. Moving coil galvanometer:– This is based on the principle that when a current carrying coil is placed in a magnetic field it experiences a torque. There is a restoring torque due to the phosphor bronze strip which brings back the coil to its normal position. In equilibrium, Deflecting torque = Restoring torque nIBA = k  [k = restoring torque/unit twist of the phosphor bronze strip] k k I   G where G  = Galvanometer constant nBA nBA  I Current sensitivity of the galvanometer is the deflection produced when unit current is passed through the galvanometer.  nBA Is   I k Voltage sensitivity is defined as the deflection produced when unit potential difference is applied across the galvanometer.   nBA Vs    [R = Resistance of the galvanometer] V IR kR  MOVING CHARGES AND MAGNETISM 15. Condition for the maximum sensitivity of the galvanometer:- The galvanometer is said to be sensitive if a small current produces a large deflection. nBA   I k   will be large if (i) n is large, (ii) B is large (iii) A is large and (iv) k is small. 16. Conversion of galvanometer into voltmeter and ammeter (a) A galvanometer is converted to voltmeter by putting a high resistance in series with it. Tot al r esist an ce of volt m et er = Rg + R where Rg is the galvonometer resistance. R is the resistance added in series. V Current through the galvanometer = Ig = Rg  R where V is the potential difference across the voltmeter. Ig Rg HR G Voltmeter I R I M N V  R= G Ig Range of the voltmeter: 0 – V volt. (b) A galvanometer is converted into an ammeter by connecting a low resistance in parallel with it (shunt)  I g  Shunt = S    R g where Rg is the galvanometere resistance.  I  Ig  Ig Rg G R I I S M (I - Ig) N I is the total current through the ammeter. Ig is the current through the ammeter. Effective resistance of the ammeter Rg R = R S g The range of the ammeter is 0 – I A. An ideal ammeter has zero resistance.